Jennifer over at Cocktail Party Physics has a nice post about her trip to Disneyland. The one ride that would be fun to play with (in terms of physics) would be the tower of terror. Think of the cool things you could do with a video camera during that ride. It would be like a mini-vomit-comet. Anyway, I want to talk about one part of Jennifer’s post.

“As one would expect, this lifted us out of our seats slightly, as much as the straps would allow, and we got that one glorious moment of seeming weightlessness, before reaching a jerky stop and being raised back up for another drop.”

It is indeed true that you would be “lifted” out of your seats, but I am not sure this is what you would expect. If the elevator is free falling, wouldn’t you just float in your seat but not be lifted up? First, I am not going to talk about weightlessness. I think I have already done that extensively in this previous post on weightlessness and apparent weight. So, why are you “lifted out of your seat”? Let me start by assuming you are sitting in your seat at rest right before you drop. In this case, you would have the following free body diagram.

I know what you are thinking – big deal. Yes, this is a simple free body diagram where the two forces have the same magnitude and the total force is zero vector. But there is a very important point. How does the chair ‘know’ exactly how much force to exert on the person? I was looking for a link, it seems like I have not blogged about this before. Wha? Ok, let me step back and draw a couple more free body diagrams. Here is a book sitting on a table (essentially the same as the the person sitting on a chair, but oh).

Suppose in this case, the box has a weight of 2 Newtons. Then obviously the table must push up on the box with 2 Newtons of force. What if the table pushed up with 3 Newtons? What if the table pushed up with 1 Newton? No, the table must push with EXACTLY 2 Newton to make the net force the zero vector. Now suppose I push down with my hand on the box with a force of 1 Newtons. Here is the free body diagram.

So, now the table has to push up with 3 Newtons. Clearly, there is something special about this table. It knows just how much force to exert on an object to make it not accelerate (and thus stay stationary). Actually, the table is not magic. It’s just made of springs (sort of). Here is a model:

In this model, the table (and the box) are made of particles connected by springs. Springs are great. When you compress them, they exert a force. The more you compress them, the greater the force. This relationship between force and compression for a spring is known as Hooke’s Law and can be written as:

So, this is how the table “knows” how hard to push up on the box. If I push down with my hand, the table gets compressed more and therefore pushes up even harder. You can actually see this happen. Take a laser pointer and shine it on a mirror or something like that on a flat table. Notice what happens to the reflected beam when you sit on the table.

Back to the falling elevator and the person. What happens when the whole room has an acceleration downward the same as a free falling object? You could think of this in the reference frame of the elevator as having no gravity (weightless). BUT, you would still be compressing the seat. There would still (momentarily) be an upward force on the person. This is what pushes the person up.

Comments

The question of how a solid knows how much normal force to exert is certainly an interesting one, but I’m not sure about the conclusion regarding people being lifted out of their ride as Disneyworld. The time for the table to deform back to its equilibrium state when you take weight off it should be something comparable to the size of the deformed region divided by the speed of sound in the table. The speed of sound in most solids is on the order of a thousands of meters per second, so I’d expect any excess upward force from the seat on a rider would last much too short a time to push them up out of the seat any significant way. This effect would, I suspect, be dwarfed by the deviation from true free fall that the ride experiences due to various friction sources.

Is it possible the ride actively drags itself down slightly faster than free fall? Alternatively, could people be subconsciously pushing down on the floor slightly? If they did that under normal circumstances, nothing would happen except the weight on their butt would go down a tiny bit. But if they were in free fall, they might be pushed up out of their seats by a small force they were exerting on the floor with their feet.

Okay, that wasn’t a very coherent comment: I hit the “Submit Comment” button before I was done rewriting.

I’ve been on the Tower of Terror at Disney parks in both Anaheim and Orlando, and there is definitely an upward “lift” for the majority of the duration of the drop, indicating that the downward acceleration of the elevator is greater than 1G. This is confirmed by other sources as well.

Thanks for the link. They may very well accelerate downward greater than 9.8 m/s^2, but I was confused by that site.

Here is a quote:

the drop. First of all it is not a drop. It forces you down. So once you leave the 5th dimension the elevator pulls on to a platform and locks into place. The platform is hanging by cables in a pully system

So, they say it is not a drop and they are talking about stuff, then they add in the 5th dimension. Weird.

@meichenl, you may well be right that the table is pushing too briefly to make a difference (I’m no physicist), but your buttocks are deforming like springs too, and they’re definitely loose and bouncy enough to push you out. Even more so if the person is apprehensive and gives them a good frightened clench as they feel the weight disappear.

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